i have two possible forms of a URL string
我有两种可能的URL字符串形式
http://www.abcexample.com/landpage/?pps=[Y/lyPw==;id_1][Y/lyP2ZZYxi==;id_2];[5403;ord];
http://www.abcexample.com/landpage/?pps=Y/lyPw==;id_1;unknown;ord;
I want to get out the Y/lyPw==
in both examples
我想在两个例子中得出Y / lyPw ==
so everything before ;id_1
between the brackets
所以之前的一切;在括号之间的id_1
will always come after the ?pps=
part
总是会出现在?pps = part之后
What is the best way to approach this? I want to use the big query language as this is where my data sits
解决这个问题的最佳方法是什么?我想使用大查询语言,因为这是我的数据所在
3 个解决方案
#1
7
Here is one way to build a regular expression to do it:
以下是构建正则表达式的一种方法:
SELECT REGEXP_EXTRACT(url, r'\?pps=;[\[]?([^;]*);') FROM
(SELECT "http://www.abcexample.com/landpage/?pps=;[XYZXYZ;id_1][XYZZZZ;id_2];[5403;ord];"
AS url),
(SELECT "http://www.abcexample.com/landpage/?pps=;XYZXYZ;id_1;unknown;ord;"
AS url)
#2
6
You can use this regex:
你可以使用这个正则表达式:
pps=\[?([^;]+)
The idea behind this regex is:
这个正则表达式背后的想法是:
pps= -> Look for the pps= pattern
\[? -> might have a [ or not
([^;]+) -> store the content up to the first semi colon
So, for your both url this regex will match (in blue) and capture (in green) as below:
所以,对于你的两个url,这个正则表达式将匹配(蓝色)和捕获(绿色),如下所示:
For BigQuery you have to use
对于BigQuery,你必须使用
REGEXP_EXTRACT('str', 'reg_exp')
Quoting its documentation:
引用其文档:
REGEXP_EXTRACT: Returns the portion of str that matches the capturing group within the regular expression.
REGEXP_EXTRACT:返回与正则表达式中的捕获组匹配的str部分。
You have to use a code like this:
你必须使用这样的代码:
SELECT
REGEXP_EXTRACT(word,r'pps=\[?([^;]+)') AS fragment
FROM
...
For a working example code you can use:
对于可以使用的工作示例代码:
SELECT
REGEXP_EXTRACT(url,r'pps=\[?([^;]+)') AS fragment
FROM
(SELECT "http://www.abcexample.com/landpage/?pps=;[XYZXYZ;id_1][XYZZZZ;id_2];[5403;ord];"
AS url),
(SELECT "http://www.abcexample.com/landpage/?pps=;XYZXYZ;id_1;unknown;ord;"
AS url)
#3
2
This regex should work for you
这个正则表达式应该适合你
(\w+);id_1
It will extract XYZXYZ
它将提取XYZXYZ
It uses the concept of Group capture
它使用了组捕获的概念
看这个演示
#1
7
Here is one way to build a regular expression to do it:
以下是构建正则表达式的一种方法:
SELECT REGEXP_EXTRACT(url, r'\?pps=;[\[]?([^;]*);') FROM
(SELECT "http://www.abcexample.com/landpage/?pps=;[XYZXYZ;id_1][XYZZZZ;id_2];[5403;ord];"
AS url),
(SELECT "http://www.abcexample.com/landpage/?pps=;XYZXYZ;id_1;unknown;ord;"
AS url)
#2
6
You can use this regex:
你可以使用这个正则表达式:
pps=\[?([^;]+)
The idea behind this regex is:
这个正则表达式背后的想法是:
pps= -> Look for the pps= pattern
\[? -> might have a [ or not
([^;]+) -> store the content up to the first semi colon
So, for your both url this regex will match (in blue) and capture (in green) as below:
所以,对于你的两个url,这个正则表达式将匹配(蓝色)和捕获(绿色),如下所示:
For BigQuery you have to use
对于BigQuery,你必须使用
REGEXP_EXTRACT('str', 'reg_exp')
Quoting its documentation:
引用其文档:
REGEXP_EXTRACT: Returns the portion of str that matches the capturing group within the regular expression.
REGEXP_EXTRACT:返回与正则表达式中的捕获组匹配的str部分。
You have to use a code like this:
你必须使用这样的代码:
SELECT
REGEXP_EXTRACT(word,r'pps=\[?([^;]+)') AS fragment
FROM
...
For a working example code you can use:
对于可以使用的工作示例代码:
SELECT
REGEXP_EXTRACT(url,r'pps=\[?([^;]+)') AS fragment
FROM
(SELECT "http://www.abcexample.com/landpage/?pps=;[XYZXYZ;id_1][XYZZZZ;id_2];[5403;ord];"
AS url),
(SELECT "http://www.abcexample.com/landpage/?pps=;XYZXYZ;id_1;unknown;ord;"
AS url)
#3
2
This regex should work for you
这个正则表达式应该适合你
(\w+);id_1
It will extract XYZXYZ
它将提取XYZXYZ
It uses the concept of Group capture
它使用了组捕获的概念
看这个演示